# A bilinear form is nonsingular and a self-adjoint operator is nonsingular.

I am reviewing some more advanced linear algebra before the semester begins, and I came accross these two proofs, and I was hoping some one could verify that my proofs are correct.

(1) The matrix, relative to any basis, of any positive definite Hermitian form is nonsingular.

(2) For any complex $$n\times n$$ matrix $$I+A^*A$$ is nonsingular, where $$A^*$$ is the conjugate transpose of $$A$$.

Here are my proofs.

(1) Let $$\beta$$ be any basis for the $$n$$ dimensional vector space in question. Let $$H$$ be any positive definite Hermitian form. Let $$A$$ be the matrix associated with $$H$$ with respect to the basis $$B$$. By that I mean $$A_{ij} = H(v_i,v_j)$$ for $$v_i,v_j\in \beta$$. Suppose that $$A$$ is singular, then there is a nonzero vector $$x$$ such that $$Ax=0$$. This implies that $$x^*Ax=0$$. Hence $$H(x,x)=x^{*}Hx=0$$. But this contradicts the fact that $$H$$ is a positive definite.

(2) Since $$(I+A^*A)^*=I^*+A^*(A^*)^*=I+A^*A$$, then the operator is $$(I+A^*A)$$ is self-adjoint. A matrix is Self-adjoint if and only if there exists a basis such that the representation of $$A$$ is diagonal. A diagonal matrix is nonsigular, which implies that $$A$$ is nonsingular.

Is there a more direct way to show (2) without using the fact that the operator is self-adjoint?

• It is worth pointing out that an arbitrary self-adjoint matrix need not be invertible (take the zero matrix, for example). – cmk Aug 15 at 23:59

If $$(I+A^{*}A)x=0$$ then $$x^{*}(I+A^{*}A)x=0$$. This can be written as $$x^{*}x+((Ax)^{*}(Ax)=0$$. Since both terms are non-negative this implies that the first term is $$0$$ and hence $$x=0$$. Thus the kernel of $$I+A^{*}A$$ is $$\{0\}$$ so it is invertible.
Similarly, for the first part $$Ax=0$$ implies $$x^{*}Ax=0$$ which implies $$x=0$$ so $$A$$ is invertible.